similar to: Variance-covariance matrix for parameter estimates

Forgive me if this is a trivial question, but I couldn't find it an answer in former forums. I'm trying to reproduce some SAS results where they set two parameters equal. For example: y = b1X1 + b2X2 + b1X3 Notice that the variables X1 and X3 both have the same slope and the intercept has been removed. How do I get an estimate of this regression model? I know how to remove the intercept

Hi all, Can you please tell me what is the problem here. My regression eq is y = B0 + B1X1 +B2X2 +e And i am interested in coefficient B1 I am doing regression with two cases: 1) reg<-lm(y ~ X1 + X2, sam) where sam is the data 2) reg<-lm(y ~ X1 + X2, sam, na.action= na.exclude) . I have missing values in X1 but the values of coefficient is not consistent in two cases. Actually B1 in

Hi everyone: ¡¡I am new to R and I have a really basic question. I have already got a generalized linear model from some dataset, say y=b0 + b1X1 + b2X2. Then I want to get the value of y provided, say, X1=1, X2=2. And the confidence Intervals of this y. I know I can just calculate that since I know the model already. But is there some code that can calculate those automatically?

Hi everyone, I'm trying to perform a linear regression y = b1x1 + b2x2 + b3x3 + b4x4 + b5x5 while constraining the coefficients such that -3 <= bi <= 3, and the sum of bi =1. I've searched R-help and have found solutions for constrained regression using quadratic programming (solve.QP) where the coefficients are between 0 and 1 and sum to 1, but unfortunately do not understand

Good day R-listers, My question is more a statistical question than an R related question, so please bear with me i'm currently applying the chow test of poolability in fact i'm working with panel N=17 T=5 , and my model looks like this : Yit= a0+B1X1+B2X2+B3X3+B4X4+eit My question is the following when i'm Testing for the equality of the coefficients of the unpooled data (the

Hi all. In a GLM in which g(mu) = b0 + b1X1 + b2X2 + b3X3 + b4X4 + b5X5 + b6X6 + b7X7, if I want to test if b1 + b5 = b2 + b6, I can use the contrast package or multicomp package. How can I do a similar test if I am fitting a GAMLSS using the gamlss package? Thank you for your help. Gustavo

Dear all, I would like to draw three forest plots to represent results at years 1, 2 and 3. I have the data as point estimates and 95% confidence intervals. Using the following code I can get three basic forest plots - the first which has the table of results. I have to plot each separately as the usual par(mfrow=c(3,1)) does not work with the function forestplot within rmeta. I can easily put

Hi! I would like to do Cox regression using the available routines in the survival package BUT I want to use an arbitrary link function, i.e. want to use the model h(t)=h_0(t)r(beta'z) with arbitrary function r, instead of h(t)=h_0(t)exp(beta'z) Grateful for any comment on this, Dragi ----------------------------------------------------------- Dragi Anevski, PhD Mathematical

Hi, In the following results I interpret exp(coef) as the factor that multiplies the base hazard rate if the corresponding variable is TRUE. For example, when the bucket is ks008 and fidelity <= 3, then the rate, compared to the base rate h_0(t), is h(t) = 0.200 h_0(t). My question is then, to what case does the base hazard rate correspond to? I would expect the reference to be the first

I have a logit model with about 10 predictors and I am trying to plot the probability curve for the model. Y=1 = 1 / 1+e^-z where z=B0 + B1X1 + ... + BnXi If the model had only one predictor, I know to do something like below. mod1 = glm(factor(won) ~ as.numeric(bid), data=mydat, family=binomial(link="logit")) all.x <- expand.grid(won=unique(won), bid=unique(bid)) y.hat.new

My question is, how to compute hazard function(H(t)) after building the coxph model. I even aware of the terminology that differs from hazard function(H(t)) and the hazard rate(h(t)). Here onward I wish to calculate both. Here what I have done in two different methods; ##########################################################################################

Hello, I am trying to learn how to perform Multiple Regression Analysis in R. I decided to take a simple example given in this PDF: http://www.utdallas.edu/~herve/abdi-prc-pretty.pdf I created a small CSV called, students.csv that contains the following data: s1 14 4 1 s2 23 4 2 s3 30 7 2 s4 50 7 4 s5 39 10 3 s6 67 10 6 Col headers: Student id, Memory span(Y), age(X1), speech rate(X2) Now

Hi I have an asterisk box with digium hardware connected to a Siemens EWSD version 15 using a crossed E1 cable. The asterisk is serving as a h323 media gateway. When i start asterisk, the Siemens gives me this alarm: REPORTES GENERADOS EN LA EWSD AES/V15SBOL/BOLCBK1V51327079/013 05-06-06 11:25:38 7773 3062/03728 HF.ARCHIVE-80040

Hi, I'm using the ks.test() on two vectors. I looked up the reference and also coded up a version of the two sample Smirnov test. My question is that how can I decide from the output of R that the two vectors x & y come from the same distribution? Am I correct in assuming that smaller D values indicate that they come from the same distribution? In addition how can I use the p value that

Is the following an inconsistency, programming glitch or a feature? One would expect that vcov(obj) was the variance-covariance of coef(obj), but apparently this is not the case for polr objects: > x <- rnorm( 100 ) > y <- rnorm( 100 ) > ff <- factor( sample( 1:4, 100, replace=T ) ) > pm <- polr( ff ~ x + y ) > coef( pm ) x y 0.21219010 0.03558506

Hi Everyone, I have data in a long format e.g. there is one row per patient but each follow-up appointment is included in the row. So, a snippet of the data looks like this: TrialNo Drug Sex Rand Adate1 Date1 Dose1 Time1 Adate2 Date2 Dose2 Time2 B1001029 LTG M 15719 30/04/2003 15825 150 106 29/08/2003 15946 200 227 B1117003 LTG M 15734 30/04/2003 15825 200 91 03/09/2003 15951 250 217

Hello again, I m trying to use timereg package as you suggested (R2.7.1 on XP Pro). here is my script based on the example from timereg for a fine & gray model in which relt = time to event, rels = status 0/1/2 2=competing, 1=event of interest, 0=censored random = covariate I want to test library(timereg) rel<-read.csv("relapse2.csv", header = TRUE, sep = ",",

Dear R-helpers, I was testing the truncgof CRAN package, found something that looked like a bug, and did my job: contacted the maintainer. But he did not reply, so I am resending my query here. I installed package truncgof and run the example for function ad.test. I got the following output: set.seed(123) treshold <- 10 xc <- rlnorm(100, 2, 2) # complete sample xt <- xc[xc >=

I don't know from the nortest package, but it should ***always*** be the case that you test hypotheses H_0: The data have a normal distribution. vs. H_a: The data do not have a normal distribution. So if you get a p-value < 0.05 you can say that ***there is evidence*** (at the 0.05 significance level) that the data are not from a normal distribution. If the nortest package does

Dear R help forum, I am using the function 'coxph' to obtain hazard ratios for the comparison of a standard treatment to new treatments. This is easily obtained by fitting the relevant model and then calling exp(coef(fit1)) say. I now want to obtain the hazard ratio for the comparison of two non-standard treatments. >From a statistical point of view, this can be achieved by dividing